COMEDK Exam 2014 Solved Paper

Let coordinate of D is (x,y,z).
Using mid-point formula,
Q=(

x−6

2

,

y−9

2

,

z+0

2

)
Also. P=(

2+6

2

,

3+9

2

,

0+0

2

)=(4,6,0)
Since, AC||PQ
∴D.R′.s of line AC=D.R′s of line PQ
⇒(−8,−12,0)=(

x−14

2

,

y−21

2

,

z

2

)
⇒x=−2,y=−3,z=0
⇒D(−2,−3,0)⇒Q(−4,−6,0)
If L is mid - point of PQ, then
L(

4−4

2

,

6−6

2

,0)=(0,0,0)
∴ Perpendicular distance of L(0,0,0) from the plane
3x+4z+25=0 is
|

3(0)+4(0)+25

√(3)2+(4)2

|=|

25

√25

|=√25=5